Rotational Motion Part 3

Concept:Conservation of mechanical energy is used. The gravitational potential energy lost by the centre of mass (COM) converts completely into rotational kinetic energy about the pivot.Explanation:First, find the moment of inertia about the pivot using the parallel axis theorem.The COM is at the centre of the rod, a distance $L/2$ from one end. The pivot is at $L/3$ from the same end.So, distance from COM to pivot is $d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$.Moment of inertia about COM: $I_{\text{cm}} = \frac{ML^2}{12}$.Thus, $I = I_{\text{cm}} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{6}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.When the rod falls from vertical to horizontal:Initial height of COM above the pivot is $h_i = \frac{L}{6}$ (since COM is above pivot).Final height of COM relative to pivot is zero (COM lies at same horizontal level as pivot when rod is horizontal on table).Loss in gravitational potential energy: $\Delta U = M g \Delta h = M g \left(\frac{L}{6}\right)$.This loss equals rotational kinetic energy at impact: $\frac{1}{2} I \omega^2$.So, $M g \frac{L}{6} = \frac{1}{2} \cdot \frac{ML^2}{9} \cdot \omega^2$.Cancel $M$ and simplify: $\frac{gL}{6} = \frac{L^2 \omega^2}{18}$ ⇒ $\frac{g}{6} = \frac{L \omega^2}{18}$ ⇒ $\omega^2 = \frac{3g}{L}$.Therefore, $\omega = \sqrt{\frac{3g}{L}}$.Answer:$\omega = \sqrt{\frac{3g}{L}}$ corresponds to option B.